安振平老师的5898号不等式问题的证明

题目:已知$a,b,c\geq 1$, 求证:$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}\leq 1+\frac{2}{1+ab+bc+ca}.$

证明:原不等式等价于

$2\,abc \left( {a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca \right) +{b}^{2}{c}^{2}+{c}^{2}{a}^{2}+{a}^{2}{b}^{2}-{a}^{3}{b}^{3}-{b}^{3}{c}^{3}-{c}^{3}{a}^{3}+3\,{a}^{2}{b}^{2}{c}^{2}$

$-abc [ bc \left( b+c \right) +ca\left( c+a \right) +ab \left( a+b \right) ] +{a}^{2}{b}^{2}{c}^{2} \left( ab+bc+ca \right)\geq 0.$                                                (1)

由$a,b,c\geq 1$，可设$a=1+x,b=1+y,c=1+z(x,y,z\geq 0)$,则不等式(1)等价于

$2\,{x}^{3}{y}^{3}z+8\,x{y}^{3}{z}^{2}+8\,x{y}^{2}{z}^{3}+8\,{x}^{2}y{z}^{3}+8\,{x}^{3}y{z}^{2}+8\,{x}^{3}{y}^{2}z+8\,{x}^{2}{y}^{3}z+6\,{x}^{3}{y}^{2}{z}^{2}$

$+6\,{x}^{2}{y}^{2}{z}^{3}+2\,{x}^{3}y{z}^{3}+6\,{x}^{2}{y}^{3}{z}^{2}+2\,x{y}^{3}{z}^{3}+{x}^{3}{y}^{3}{z}^{2}+{x}^{2}{y}^{3}{z}^{3}+{x}^{3}{y}^{2}{z}^{3}$

$+34\,{x}^{2}y{z}^{2}+34\,x{y}^{2}{z}^{2}+37\,{y}^{2}zx+10\,xy{z}^{3}+37\,yz{x}^{2}+10\,x{y}^{3}z+36\,xyz+10\,{x}^{3}yz+30\,{x}^{2}{y}^{2}{z}^{2}$

$+37\,y{z}^{2}x+34\,{x}^{2}{y}^{2}z+4\,z{x}^{2}+4\,z{y}^{2}+4\,y{z}^{2}+4\,x{z}^{2}+4\,y{x}^{2}+4\,x{y}^{2}+{y}^{3}z+y{z}^{3}+2\,{y}^{2}{z}^{2}$

$+{x}^{3}z+x{z}^{3}+2\,{x}^{2}{z}^{2}+{x}^{3}y+x{y}^{3}+2\,{x}^{2}{y}^{2}+4\,xy+4\,yz+4\,xz\geq 0.$                                                                                                   (2)

不等式(2)显然成立，故原不等式获证.